Constant acceleration trajectory

In summary, the conversation discusses a problem where a point body is thrown in space from one point to another, with given distances, angles, and velocities. The problem can be solved by decomposing the velocity and acceleration vectors into components and using the equation s = v0i t + 1/2 ai t^2. Another approach is to consider the trajectory as a parabola and find points on it with a given distance and slope.
  • #1
jal3ous
5
0
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Hello,

I'v been trying to find an answer for this question for quite some time, I always hit a complex trigonometric equation or a 4th degree polynomial equation. any help would be very much appreciated!
Here's the problem:

a point body (B) is thrown in space (no friction) from a point (P1) with an initial velocity vector (V) and a constant acceleration vector (A). The trajectory generated by (B) passes through a point (P2).

We know the distance between (P1) and (P2) = d
We know the angle between the vector (P1P2) and the vector (V) = alpha
We know the magnitude of (V) = v
We know the magnitude of (A) = a

Calculate the angle (theta) between (V) and (A)
 
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  • #2
Have you tried to decompose v and a into components parallel and perpendicular to P1P2, and then using s = v0i t + 1/2 ai t^2 for each of them? The time t is equal in both equations. while s = 0 for perpendicular and s = d for parallel.
 
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  • #3
Yes, I started by doing that. I ended up with a 4th degree polynomial with long and complex symbolic coefficients. I checked up the general form of the solution for 4th degree equation and it's huge even without long and complex symbolic coefficients. isn't there any easyer way to solve it? Thank you :)
 
  • #4
I think the problem is somehow equivalent to finding the slope of a field given the slant range and the initial velocity vector of a projectite, but I couldn't go any further from there...
 
  • #5
jal3ous said:
isn't there any easyer way to solve it?
An idea, not sure if simpler :

The trajectory is a parabola (lets assume y-axis is along a). You look for two points on it, with a given distance, and parabola slope at one of them related to the direct slope between them.
 

Related to Constant acceleration trajectory

What is a constant acceleration trajectory?

A constant acceleration trajectory is a type of motion where the acceleration of an object remains constant throughout its entire path. This means that the object's velocity changes at a constant rate, resulting in a linear change in its position over time.

How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken for the change in velocity to occur. This formula is also known as the "equation of motion" or "kinematic equation" for constant acceleration.

What are some real-world examples of constant acceleration?

Some common examples of constant acceleration include a car accelerating from a stop, a skydiver falling towards the ground, and a rocket launching into space. These objects experience a constant force, resulting in a constant acceleration throughout their motion.

What is the difference between constant acceleration and uniform motion?

Constant acceleration and uniform motion are two different types of motion. In constant acceleration, the acceleration remains the same, resulting in a changing velocity. In uniform motion, the velocity remains constant, resulting in no acceleration. In other words, constant acceleration is a type of accelerated motion, while uniform motion is a type of unaccelerated motion.

How is constant acceleration related to Newton's Laws of Motion?

Constant acceleration is directly related to Newton's Second Law of Motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In the case of constant acceleration, the force remains constant, resulting in a linear change in the object's velocity over time.

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